An urn contains $4$ red marbles and $8$ black marbles. What is the probability that exactly $2$ red marbles are drawn before the third black marble if the marbles are drawn without replacement?
I happened to get the correct answer, which is:
$\left(\frac{4}{12}\right)\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\left(\frac{7}{9}\right)\left(\frac{6}{8}\right)3!$
but somehow I do not understand why we should multiply all the numbers with $3!$
Can someone explain to me?
Thank you!
Equivalently, the first five marbles are two reds and three blacks and these come in an order such that the last one is black.
Out of the $12\choose 5$ ways to pick $5$ marbles, there are ${4\choose 2}{8\choose 3}$ ways to pick $2$ reds and $3$ blacks. So the probability of drawing two reds and three blacks (in any order) is $$ \frac{{4\choose 2}{8\choose 3}}{12\choose 5}=\frac{14}{33}.$$ However, the probability that the last of such five marbles is one of the three blacks, is $\frac 35$. Hence we have to multiply by that: $$ \frac{14}{33}\cdot\frac35=\frac{14}{55}.$$